# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4

Question 1.
$$\frac { { e }^{ x } }{ sinx }$$
Solution:
$$y=\frac { { e }^{ x } }{ sinx }$$
$$for\quad y=\frac { u }{ v } ,$$
$$\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }$$
$$or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z$$

Question 2.
$${ e }^{ { sin }^{ -1 }x }$$
Solution:

Question 3.
$${ e }^{ { x }^{ 3 } }$$ = y
Solution:
Let y = $${ e }^{ { x }^{ 3 } }$$
Differentiating w.r.t. x,
$$\frac{d y}{d x}=\frac{d}{d x}\left(e^{x^{3}}\right)$$ = $$e^{x^{3}} \cdot \frac{d}{d x}\left(x^{3}\right)$$
= $${ e }^{ { x }^{ 3 } }$$.3x² = 3x²$${ e }^{ { x }^{ 3 } }$$

Question 4.
$$sin\left( { tan }^{ -1 }{ e }^{ -x } \right)$$ = y
Solution:
$$sin\left( { tan }^{ -1 }{ e }^{ -x } \right)$$ = y
$$\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right)$$
$$=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } \frac { d }{ dx } \left( { e }^{ -x } \right)$$
$$=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right)$$

Question 5.
$$log(cos\quad { e }^{ x })$$ = y
Solution:
$$\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right)$$

Question 6.
$${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$$…$$+{ e }^{ { x }^{ 5 } }$$ = y(say)
Solution:

Question 7.
$$\sqrt { { e }^{ \sqrt { x } } }$$, x > 0
Solution:

Question 8.
log(log x), x > 1
Solution:
y = log(log x), x > 1
Differentiating w.r.t. x,
$$\frac{d y}{d x}$$ = $$\frac{1}{\log x} \cdot \frac{d}{d x}$$(log x)
= $$\frac{1}{\log x} \cdot \frac{1}{x}$$ = $$\frac{1}{x \log x}$$, x > 1

Question 9.
$$\frac { cosx }{ logx }$$ = y(say),x>0
Solution:

Question 10.
cos(log x + ex), x > 0
Solution:

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